Question Number 191529 by MATHEMATICSAM last updated on 25/Apr/23 $$\mathrm{If}\:\frac{{x}\:−\:{y}}{{x}\sqrt{{y}}\:+\:{y}\sqrt{{x}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{{x}}}\:;\:\left({x}\:>\:\mathrm{0}\:\mathrm{and}\:{y}\:>\:\mathrm{0}\right)\:\mathrm{then} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\frac{{x}}{{y}}\:. \\ $$ Answered by mehdee42 last updated on 25/Apr/23 $$\frac{\left(\sqrt{{x}}−\sqrt{{y}}\right)\left(\sqrt{{x}}+\sqrt{{y}}\right)}{\:\sqrt{{xy}}\left(\sqrt{{x}}+\sqrt{{y}}\right)}=\frac{\mathrm{1}}{\:\sqrt{{x}}} \\ $$$${x}−\sqrt{{xy}}=\sqrt{{xy}}\Rightarrow{x}=\mathrm{2}\sqrt{{xy}}\Rightarrow\frac{{x}}{{y}}=\mathrm{4}\:\checkmark \\…
Question Number 191530 by Shrinava last updated on 25/Apr/23 Commented by mr W last updated on 25/Apr/23 $${question}\:{is}\:{wrong}.\:{figure}\:{is}\:{not} \\ $$$${possible}.\:{y}=\mathrm{0}. \\ $$ Commented by Shrinava…
Question Number 191525 by mehdee42 last updated on 25/Apr/23 $${Q}\::\:{Show}\:{that}\:{the}\:{numbers}\:\sqrt{\mathrm{3}\:}\:,\:\mathrm{2}\:\&\:\sqrt{\mathrm{8}}\:{cannot}\:{be}\:{terms}\:{of}\:{an}\:{arithmetic}\:{sequence}. \\ $$ Answered by mr W last updated on 26/Apr/23 $$\sqrt{\mathrm{3}}<\mathrm{2}<\sqrt{\mathrm{8}} \\ $$$${assume}\:{they}\:{are}\:{terms}\:{of}\:{an}\:{A}.{P}., \\ $$$${then}\:{we}\:{have}…
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Question Number 191527 by MATHEMATICSAM last updated on 25/Apr/23 $${x}\:+\:{y}\:=\:\mathrm{1}\:\mathrm{and}\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:=\:\mathrm{2}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:{x}^{\mathrm{11}} \:+\:{y}^{\mathrm{11}} . \\ $$ Answered by Rasheed.Sindhi last updated on 25/Apr/23…
Question Number 60445 by ANTARES VY last updated on 21/May/19 Commented by bhanukumarb2@gmail.com last updated on 21/May/19 $${cauchy}\:{inequality}\: \\ $$ Answered by MJS last updated…
Question Number 191512 by mnjuly1970 last updated on 25/Apr/23 $$ \\ $$$$\:\:\:\:\:{Q}:\:\:\:\:\:\:\:{the}\:{equation}\: \\ $$$$\:\:\:\: \\ $$$$\:\:\:\lfloor\:\mathrm{cos}\left(\mathrm{4}{x}\:\right)\rfloor={m}.\mathrm{cos}\left(\mathrm{2}{x}\right) \\ $$$$\:\:\:{has}\:{no}\:\:{solution}\:.\:\:{x}\in\:\left(\mathrm{0},\:\frac{\pi}{\mathrm{2}}\:\right) \\ $$$$\:\:\:\:{find}\:{the}\:{acceptable} \\ $$$$\:\:\:\:\:\:\:{real}\:{values}\:{for}\:\:\:\:''{m}''. \\ $$ Answered…
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Question Number 125960 by AST last updated on 26/Sep/22 $$ \\ $$ Answered by mahdipoor last updated on 16/Dec/20 $$\begin{cases}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mid{a}\mid^{\mathrm{2}} +\mid{b}\mid^{\mathrm{2}} }\\{\mathrm{2}{ab}\leqslant\mathrm{2}\mid{a}\mid.\mid{b}\mid}\end{cases} \\…